Column Force
Column Force
(OP)
We have some new calculation standards in Europe for
steel and I want ask You ALL about column calculation.
If we have a column RHS 100x100x4 ( RHS 3.937x3.937x 0.15748 inch )
Length = L = 6000mm= 236 " Ix=Iy=2263500 mm4 E= 210000 N/mm2
Material normal structural carbon steel .
If both ends are pinned as Euler 2 condition, what is
acceptable maximum compression force for this column ?
The force type is static weight .
The reason for this is that I get different results with different standards !
I do NOT use results for some structure !! This is only a "test" to see differencies
for calculation methods.
Answer is = ?
Best Regards
Markku Lavi
Lamek Oy
steel and I want ask You ALL about column calculation.
If we have a column RHS 100x100x4 ( RHS 3.937x3.937x 0.15748 inch )
Length = L = 6000mm= 236 " Ix=Iy=2263500 mm4 E= 210000 N/mm2
Material normal structural carbon steel .
If both ends are pinned as Euler 2 condition, what is
acceptable maximum compression force for this column ?
The force type is static weight .
The reason for this is that I get different results with different standards !
I do NOT use results for some structure !! This is only a "test" to see differencies
for calculation methods.
Answer is = ?
Best Regards
Markku Lavi
Lamek Oy






RE: Column Force
Looking at your calculation, first of all you need to calculate the principle axes of the angle and then calculate the minimum second moment of area.
The principle axes for an equal angle run at 45 degrees to
the X and Y axes and for your particular angle the minimum
value of Ixy = 465407.96mm ^4
Now using Euler buckling load formula for both ends pinned
Pe=3.142^2 * E * I/(L^2)
Pe= 3.142^2 * 210000 * 465407.96/(6000^2)
Pe = 2714.87 N
where Pe= buckling load
E= modulus of elasticity of material
I= second moment of area
hope this helps
regards
desertfox
RE: Column Force
I assume that the section RHS 3.937x3.937x 0.15748 inch refers to a square HSS and not an angle shape (Ix=Iy=2.75E5 for a square vs. 1.14E6 for an angle. This is for a 4.78mm section). If the section is square, then.
The area is approximately 3.937*.15478*4=2.48 sq.in. and the radius of gyration is approx. 37.61 in. (sqrt(I/A)). The I for an axis through the diagonal is approximately 2.2634*1.414 (sqrt(2) or sqrt(Ix*Ix + Iy*Iy))
The kh/r is 159 so it falls into the intermediate range for compression members. This is the range where internal stresses have a determining effect on the strength.
The allowable stress for 350 MPa yield steel (common HSS stuff) is dependant on if it is stress relieved or heat treated. Cr/A is 64.5 MPa for regular 350 grade stuff and 68.7 MPa for heat treated or Type H.
If the section is an angle, then 'scrap' the above; I don't know what the European standards are... and they may vary from the above.
Dik
RE: Column Force
2.75E5 should read 2.75E6 (typo?)
37.61 in. should read 37.61 mm (carelessness)
2.2634*1.414 should read 2.2634E6*1.414(ditto)
The calculation of Cr is a little more involved for intermediate compression members; I can fax you the method of determining it since the formulae are a little unwieldly.